帕塞瓦尔定理Parseval‘s theorem

目录

一、Parseval's theorem介绍

二、 Parseval's theorem基本形式的证明

2.1连续信号

2.2离散信号 

三、MATLAB仿真验证 


一、Parseval's theorem介绍

帕塞瓦尔定理Parseval's theorem表明了信号的能量在时域和频域相等。

\int_{-\infty}^{\infty}|f(t)|^{2} \mathrm{~d} t=\frac{1}{2 \pi} \int_{-\infty}^{\infty}|F(\omega)|^{2} \mathrm{~d} \omega=\int_{-\infty}^{\infty}|\hat{F}(f)|^{2} \mathrm{~d} f

对于两个不同的信号,与上式对应,一个相似的等式为 

\int_{-\infty}^{\infty} f_{i}(t) f_{j}(t) \mathrm{d} t=\frac{1}{2 \pi} \int_{-\infty}^{\infty} F_{i}(\omega) F_{j}^{*}(\omega) \mathrm{d} \omega 

写成离散形式为: \sum_{n=0}^{N-1} x[n] y^{*}[n]=\frac{1}{N} \sum_{k=0}^{N-1} X[k] Y^{*}[k]

二、 Parseval's theorem基本形式的证明

2.1连续信号

\begin{aligned} & \int_{-\infty}^{\infty}|f(t)|^{2} d t \\ = & \int_{-\infty}^{\infty} f(t) f^{*}(t) d t \\ = & \int_{-\infty}^{\infty}\left[\frac{1}{2 \pi} \int_{-\infty}^{\infty} F(\omega) e^{j \omega t} d \omega\right]\left[\frac{1}{2 \pi} \int_{-\infty}^{\infty} F^{*}\left(\omega^{\prime}\right) e^{-j \omega^{\prime} t} d \omega^{\prime}\right] d t \\ = & \frac{1}{2 \pi} \int_{-\infty}^{\infty} F(\omega) \frac{1}{2 \pi} \int_{-\infty}^{\infty} F^{*}\left(\omega^{\prime}\right)\left[\int_{-\infty}^{\infty} e^{j\left[\omega-\omega^{\prime}\right] t} d t\right] d \omega^{\prime} d \omega \\ = & \frac{1}{2 \pi} \int_{-\infty}^{\infty} F(\omega) \frac{1}{2 \pi} \int_{-\infty}^{\infty} F^{*}\left(\omega^{\prime}\right) 2 \pi \delta\left(\omega-\omega^{\prime}\right) d \omega^{\prime} d \omega \\ = & \frac{1}{2 \pi} \int_{-\infty}^{\infty} F(\omega) F^{*}(\omega) d \omega \\ = & \frac{1}{2 \pi} \int_{-\infty}^{\infty}|F(\omega)|^{2} d \omega \end{aligned}

2.2离散信号 

\large \begin{array}{l} \sum_{n=0}^{N-1} x[n] y^{*}[n]=\frac{1}{N} \sum_{k=0}^{N-1} X[k] Y^{*}[k] \\ =\sum_{n=0}^{N-1} x[n]\left(\frac{1}{N} \sum_{k=0}^{N-1} Y[k] W_{N}^{-n k}\right)^{*} \\ =\frac{1}{N} \sum_{n=0}^{N-1} x[n]\left(\sum_{k=0}^{N-1} Y^{*}[k] W_{N}^{n k}\right) \\ =\frac{1}{N} \sum_{k=0}^{N-1} Y^{*}[k]\left(\sum_{n=0}^{N-1} x[n] W_{N}^{n k}\right) \\ =\frac{1}{N} \sum_{k=0}^{N-1} Y^{*}[k] X[k] \end{array}

三、MATLAB仿真验证 

clc,clear;
T = 20;  dt = 0.001; t = dt:dt:T; N = size(t,2);  
fs = 1/dt;   ws = fs*2*pi; df = 1/T; dw = 2*pi*df;
s1 = exp(-0.2*t).*sin(2*pi*0.5*t); s2 = exp(-0.1*t).*sin(2*pi*0.6*t);
Us1 = fft(s1); Us2 = fft(s2);
Et11 = sum(s1.*s1)
Ew11 = sum(Us1.*conj(Us1))/N

Et12 = sum(s1.*s2)
Ew12 = sum(Us1.*conj(Us2))/N

 结果如图所示,可以验证公式准确性帕塞瓦尔定理Parseval‘s theorem_第1张图片

 

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